How print all actors in akka system?

Question

I have created akka system. Suppose there are some actors in it. How I can print all actors from akka system with their path? (for debug purposes)

Answer 1

ActorSystem has private method printTree, that you can use for debugging.

1) Private method caller (from https://gist.github.com/jorgeortiz85/908035):

class PrivateMethodCaller(x: AnyRef, methodName: String) {
  def apply(_args: Any*): Any = {
    val args = _args.map(_.asInstanceOf[AnyRef])

    def _parents: Stream[Class[_]] = Stream(x.getClass) #::: _parents.map(_.getSuperclass)

    val parents = _parents.takeWhile(_ != null).toList
    val methods = parents.flatMap(_.getDeclaredMethods)
    val method = methods.find(_.getName == methodName).getOrElse(throw new IllegalArgumentException("Method " + methodName + " not found"))
    method.setAccessible(true)
    method.invoke(x, args: _*)
  }
}

class PrivateMethodExposer(x: AnyRef) {
  def apply(method: scala.Symbol): PrivateMethodCaller = new PrivateMethodCaller(x, method.name)
}

2) Usage

val res = new PrivateMethodExposer(system)('printTree)()
println(res)

Will print:

-> / LocalActorRefProvider$$anon$1 class akka.actor.LocalActorRefProvider$Guardian status=0 2 children
   ⌊-> system LocalActorRef class akka.actor.LocalActorRefProvider$SystemGuardian status=0 3 children
   |   ⌊-> deadLetterListener RepointableActorRef class akka.event.DeadLetterListener status=0 no children
   |   ⌊-> eventStreamUnsubscriber-1 RepointableActorRef class akka.event.EventStreamUnsubscriber status=0 no children
   |   ⌊-> log1-Logging$DefaultLogger RepointableActorRef class akka.event.Logging$DefaultLogger status=0 no children
   ⌊-> user LocalActorRef class akka.actor.LocalActorRefProvider$Guardian status=0 1 children
...

Beware, it can cause OOM If you have a lot of actors.

Answer 2

According to the documentation you can use ActorSelection with a wildcard * to make actors send identifying messages. You can have an actor that collect the ActorRefs.

As mentioned by @chris-martin only actors that are not currently busy will send. A very simple actor:

// make all the available actor to send an identifying message
public void freeActors()
{
  ActorSelection selection =
    getContext().actorSelection("/user/*");
  selection.tell(new Identify(identifyId), getSelf());
}

...
// collect responses
@Override
public void onReceive(Object message) {
  if (message instanceof ActorIdentity) {
    ActorIdentity identity = (ActorIdentity) message;
    // get the ref of the sender 
    ActorRef ref = identity.getRef();
    // the sender is up and available
   ...

EDIT: I know this is for Java, but it seemed helpful to me.

Answer 3

This reply by Roland Kuhn suggests that this isn't a completely trivial problem, but you can get pretty close (for actors that will reply to messages in a reasonable time) using the Identify-ActorIdentity request-response protocol that all actors obey.

Some untested code thrown together to illustrate the idea:

import akka.actor._

def receive = {
    case 'listActors =>
      context.actorSelection("/user/*") ! Identify()

    case path: ActorPath =>
      context.actorSelection(path / "*") ! Identify()

    case ActorIdentity(_, Some(ref)) =>
      log.info("Got actor " + ref.path.toString)
      self ! ref.path
}